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# Number of subarrays having sum in a given range

Given an array arr[] of positive integers and a range (L, R). Find the number of subarrays having sum in the range L to R.

Examples:

```Input : arr[] = {1, 4, 6}, L = 3, R = 8
Output : 3
The subarrays are {1, 4}, {4}, {6}.

Input : arr[] = {2, 3, 5, 8}, L = 4, R = 13
Output : 6
The subarrays are {2, 3}, {2, 3, 5}, {3, 5},
{5}, {5, 8}, {8}.```

A simple solution is to one by one consider each subarray and find its sum. If the sum lies in the range [L, R], then increment the count. The time complexity of this solution is O(n^2).

An efficient solution is to first find the number of subarrays having sum less than or equal to R. From this count, subtract the number of subarrays having sum less than or equal to L-1. To find the number of subarrays having sum less than or equal to the given value, the linear time method using the sliding window discussed in the following post is used:

Steps to solve this problem:

1. Declare a variable Rcnt and store the value of function countSub(arr,n,R)

2. Declare a variable lcnt and store the variable of function countSub(arr,n,L-1).

3. Return Rcnt-Lcnt.

*In countSub function:

1. Declare variables st=0,end=0,sum=0,and cnt=0.

2. While end is smaller than n:

*Update sum+=arr[end].

*While St is smaller than equal to n and sum is greater than x:

*Update sum-=arr[St] and increment St.

*Update cnt+=(end-st+1) and increment end.

3. Return cnt.

Below is the implementation of above approach:

## C++

 `// CPP program to find number of subarrays having``// sum in the range L to R.``#include ``using` `namespace` `std;` `// Function to find number of subarrays having``// sum less than or equal to x.``int` `countSub(``int` `arr[], ``int` `n, ``int` `x)``{` `    ``// Starting index of sliding window.``    ``int` `st = 0;` `    ``// Ending index of sliding window.``    ``int` `end = 0;` `    ``// Sum of elements currently present``    ``// in sliding window.``    ``int` `sum = 0;` `    ``// To store required number of subarrays.``    ``int` `cnt = 0;` `    ``// Increment ending index of sliding``    ``// window one step at a time.``    ``while` `(end < n) {` `        ``// Update sum of sliding window on``        ``// adding a new element.``        ``sum += arr[end];` `        ``// Increment starting index of sliding``        ``// window until sum is greater than x.``        ``while` `(st <= end && sum > x) {``            ``sum -= arr[st];``            ``st++;``        ``}` `        ``// Update count of number of subarrays.``        ``cnt += (end - st + 1);``        ``end++;``    ``}` `    ``return` `cnt;``}` `// Function to find number of subarrays having``// sum in the range L to R.``int` `findSubSumLtoR(``int` `arr[], ``int` `n, ``int` `L, ``int` `R)``{` `    ``// Number of subarrays having sum less``    ``// than or equal to R.``    ``int` `Rcnt = countSub(arr, n, R);` `    ``// Number of subarrays having sum less``    ``// than or equal to L-1.``    ``int` `Lcnt = countSub(arr, n, L - 1);` `    ``return` `Rcnt - Lcnt;``}` `// Driver code.``int` `main()``{``    ``int` `arr[] = { 1, 4, 6 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``int` `L = 3;``    ``int` `R = 8;` `    ``cout << findSubSumLtoR(arr, n, L, R);``    ``return` `0;``}`

## Java

 `// Java program to find number``// of subarrays having sum in``// the range L to R.``import` `java.io.*;` `class` `GFG``{``    ` `    ``// Function to find number``    ``// of subarrays having sum``    ``// less than or equal to x.``    ``static` `int` `countSub(``int` `arr[],``                        ``int` `n, ``int` `x)``    ``{``    ` `        ``// Starting index of``        ``// sliding window.``        ``int` `st = ``0``;``    ` `        ``// Ending index of``        ``// sliding window.``        ``int` `end = ``0``;``    ` `        ``// Sum of elements currently``        ``// present in sliding window.``        ``int` `sum = ``0``;``    ` `        ``// To store required``        ``// number of subarrays.``        ``int` `cnt = ``0``;``    ` `        ``// Increment ending index``        ``// of sliding window one``        ``// step at a time.``        ``while` `(end < n)``        ``{``    ` `            ``// Update sum of sliding``            ``// window on adding a``            ``// new element.``            ``sum += arr[end];``    ` `            ``// Increment starting index``            ``// of sliding window until``            ``// sum is greater than x.``            ``while` `(st <= end && sum > x)``            ``{``                ``sum -= arr[st];``                ``st++;``            ``}``    ` `            ``// Update count of``            ``// number of subarrays.``            ``cnt += (end - st + ``1``);``            ``end++;``        ``}``    ` `        ``return` `cnt;``    ``}``    ` `    ``// Function to find number``    ``// of subarrays having sum``    ``// in the range L to R.``    ``static` `int` `findSubSumLtoR(``int` `arr[], ``int` `n,``                              ``int` `L, ``int` `R)``    ``{``    ` `        ``// Number of subarrays``        ``// having sum less than``        ``// or equal to R.``        ``int` `Rcnt = countSub(arr, n, R);``    ` `        ``// Number of subarrays``        ``// having sum less than``        ``// or equal to L-1.``        ``int` `Lcnt = countSub(arr, n, L - ``1``);``    ` `        ``return` `Rcnt - Lcnt;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `arr[] = { ``1``, ``4``, ``6` `};``        ``int` `n = arr.length;``    ` `        ``int` `L = ``3``;``        ``int` `R = ``8``;``    ` `        ``System.out.println(findSubSumLtoR(arr, n, L, R));``    ``}``}` `// This code is contributed``// by Mahadev99`

## C#

 `// C# program to find number``// of subarrays having sum in``// the range L to R.``using` `System;` `class` `GFG``{``    ` `    ``// Function to find number``    ``// of subarrays having sum``    ``// less than or equal to x.``    ``static` `int` `countSub(``int``[] arr,``                        ``int` `n, ``int` `x)``    ``{``    ` `        ``// Starting index of``        ``// sliding window.``        ``int` `st = 0;``    ` `        ``// Ending index of``        ``// sliding window.``        ``int` `end = 0;``    ` `        ``// Sum of elements currently``        ``// present in sliding window.``        ``int` `sum = 0;``    ` `        ``// To store required``        ``// number of subarrays.``        ``int` `cnt = 0;``    ` `        ``// Increment ending index``        ``// of sliding window one``        ``// step at a time.``        ``while` `(end < n)``        ``{``    ` `            ``// Update sum of sliding``            ``// window on adding a``            ``// new element.``            ``sum += arr[end];``    ` `            ``// Increment starting index``            ``// of sliding window until``            ``// sum is greater than x.``            ``while` `(st <= end && sum > x)``            ``{``                ``sum -= arr[st];``                ``st++;``            ``}``    ` `            ``// Update count of``            ``// number of subarrays.``            ``cnt += (end - st + 1);``            ``end++;``        ``}``    ` `        ``return` `cnt;``    ``}``    ` `    ``// Function to find number``    ``// of subarrays having sum``    ``// in the range L to R.``    ``static` `int` `findSubSumLtoR(``int``[] arr, ``int` `n,``                              ``int` `L, ``int` `R)``    ``{``    ` `        ``// Number of subarrays``        ``// having sum less than``        ``// or equal to R.``        ``int` `Rcnt = countSub(arr, n, R);``    ` `        ``// Number of subarrays``        ``// having sum less than``        ``// or equal to L-1.``        ``int` `Lcnt = countSub(arr, n, L - 1);``    ` `        ``return` `Rcnt - Lcnt;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int``[] arr = { 1, 4, 6 };``        ``int` `n = arr.Length;``    ` `        ``int` `L = 3;``        ``int` `R = 8;``    ` `        ``Console.Write(findSubSumLtoR(arr, n, L, R));``    ``}``}` `// This code is contributed``// by ChitraNayal`

## Python 3

 `# Python 3 program to find``# number of subarrays having``# sum in the range L to R.` `# Function to find number``# of subarrays having sum``# less than or equal to x.``def` `countSub(arr, n, x):``    ` `    ``# Starting index of``    ``# sliding window.``    ``st ``=` `0` `    ``# Ending index of``    ``# sliding window.``    ``end ``=` `0` `    ``# Sum of elements currently``    ``# present in sliding window.``    ``sum` `=` `0` `    ``# To store required``    ``# number of subarrays.``    ``cnt ``=` `0` `    ``# Increment ending index``    ``# of sliding window one``    ``# step at a time.``    ``while` `end < n :``        ` `        ``# Update sum of sliding``        ``# window on adding a``        ``# new element.``        ``sum` `+``=` `arr[end]` `        ``# Increment starting index``        ``# of sliding window until``        ``# sum is greater than x.``        ``while` `(st <``=` `end ``and` `sum` `> x) :``            ``sum` `-``=` `arr[st]``            ``st ``+``=` `1` `        ``# Update count of``        ``# number of subarrays.``        ``cnt ``+``=` `(end ``-` `st ``+` `1``)``        ``end ``+``=` `1` `    ``return` `cnt` `# Function to find number``# of subarrays having sum``# in the range L to R.``def` `findSubSumLtoR(arr, n, L, R):``    ` `    ``# Number of subarrays``    ``# having sum less``    ``# than or equal to R.``    ``Rcnt ``=` `countSub(arr, n, R)` `    ``# Number of subarrays``    ``# having sum less than``    ``# or equal to L-1.``    ``Lcnt ``=` `countSub(arr, n, L ``-` `1``)` `    ``return` `Rcnt ``-` `Lcnt` `# Driver code``arr ``=` `[ ``1``, ``4``, ``6` `]``n ``=` `len``(arr)``L ``=` `3``R ``=` `8``print``(findSubSumLtoR(arr, n, L, R))` `# This code is contributed``# by ChitraNayal`

## PHP

 ` ``\$x``)``        ``{``            ``\$sum` `-= ``\$arr``[``\$st``];``            ``\$st``++;``        ``}` `        ``// Update count of``        ``// number of subarrays.``        ``\$cnt` `+= (``\$end` `- ``\$st` `+ 1);``        ``\$end``++;``    ``}` `    ``return` `\$cnt``;``}` `// Function to find number``// of subarrays having sum``// in the range L to R.``function` `findSubSumLtoR(&``\$arr``, ``\$n``, ``\$L``, ``\$R``)``{``    ``// Number of subarrays``    ``// having sum less``    ``// than or equal to R.``    ``\$Rcnt` `= countSub(``\$arr``, ``\$n``, ``\$R``);` `    ``// Number of subarrays``    ``// having sum less``    ``// than or equal to L-1.``    ``\$Lcnt` `= countSub(``\$arr``, ``\$n``, ``\$L` `- 1);` `    ``return` `\$Rcnt` `- ``\$Lcnt``;``}` `// Driver code.``\$arr` `= ``array``( 1, 4, 6 );``\$n` `= sizeof(``\$arr``);``\$L` `= 3;``\$R` `= 8;``echo` `findSubSumLtoR(``\$arr``, ``\$n``, ``\$L``, ``\$R``);` `// This code is contributed``// by ChitraNayal``?>`

## Javascript

 ``

Output

`3`

Complexity Analysis:

• Time Complexity: O(n)
• Auxiliary Space: O(1)

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